Rotation conservation of energy

1. The problem statement, all variables and given/known data

Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R. The cylinder rotates without friction about a horizontal axle along the cylinder axis. One end of the rope is attached to the cylinder. The cylinder starts with angular speed wo. After one revolution of the cylinder the rope has unwrapped and, at this instant, hangs vertically down, tangent to the cylinder.

a.Find the angular speed of the cylinder at this time. You can ignore the thickness of the rope. (Hint: Use Equation U=mgycm.)
Express your answer in terms of the variables m, M, R, and appropriate constants.

b.Find the linear speed of the lower end of the rope at this time.
Express your answer in terms of the variables m, M, R, and appropriate constants.


2. Relevant equations

I=.5MR^2
E=Iw^2

3. The attempt at a solution
I initially set it up using the conservation of energy. mgh + .5Iwo^2=.5mv^2 + .5Iwf^2. The thing is, the answer is not dependent upon h (I even tried substituting y for h) and I don't think it it supposed to include v either. Is this equation even correct? Thanks so much for any help!
 

tiny-tim

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Hi Julie! Welcome to PF! :smile:

(have an omega: ω and try using the X2 and X2 icons just above the Reply box :wink:)
… I initially set it up using the conservation of energy. mgh + .5Iwo^2=.5mv^2 + .5Iwf^2. The thing is, the answer is not dependent upon h (I even tried substituting y for h) and I don't think it it supposed to include v either. Is this equation even correct? Thanks so much for any help!
I assume v is the speed of the (centre of the) rope? Then you can write it in terms of ω and R (and remember h is a multiple of R). :wink:

Looks correct to me. :smile:
 
Re: Rotation

okay, well I tried that and I got sqrt((8[tex]\pi[/tex]mgR+.25MR2[tex]\omega[/tex] -.5m([tex]\omega[/tex]0R)2)/MR2

It said it was wrong...do you see anywhere else I messed up? Thanks!
 
Re: Rotation

Sorry, some of the symbols messed up, but in the second term the omega is squared
 

tiny-tim

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Hi Julie! :smile:


(just got up :zzz: …)
okay, well I tried that and I got sqrt((8[tex]\pi[/tex]mgR+.25MR2[tex]\omega[/tex] -.5m([tex]\omega[/tex]0R)2)/MR2

It said it was wrong...do you see anywhere else I messed up? Thanks!
(have a pi: π and an omega: ω and a square-root: √ :smile:)

I'm not sure what this is, part of it seems to be the energy (but I think a sign is wrong, and where your 8πR come from?), but what is the rest? :confused:

(an actual equation would help me :wink:)
 
Re: Rotation

haha I am sorry.
All of that came from just solving the energy equation that I initially posted for omega. I plugged .5MR2 in for I and omega R for v. The 8[tex]\pi[/tex]R came from taking the circumference of the cylinder as 2[tex]\pi[/tex]R and plugging that in for the height, and then multiplying by 4 from solving for omega.

After thinking about it though, I probably need to divide the circumference by 2 to make the height for the center of mass of rope. Does that make sense? Unfortunately I only have one attempt left on this question, otherwise I would try and see if that was the only problem with it. Thanks!
 

tiny-tim

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Hi Julie! :smile:

(have a pi: π :wink:)
After thinking about it though, I probably need to divide the circumference by 2 to make the height for the center of mass of rope.
Yes!! :smile:

But I still don't understand where the factor of 4 came from, nor what the expression as a whole is supposed to be. :confused:

Is it suppposed to be ω? If so, it shouldn't have ω in it, should it? :redface:

I think you'd better write out your final attempt here, before you commit yourself. :wink:
 
Re: Rotation

π²³ ∞ ° → ~ µ ρ σ τ ω ∑ … √ ∫ ≤ ≥ ± ∃ · θ φ ψ Ω α β γ δ ∂ ∆ ∇ ε λ Λ Γ ô

Yes, I am solving for the final angular velocity, so all ω's in my answer are the initial angular velocity of the cylinder.

Okay, so starting with the energy equation:
mgh + .5Iω02=.5m(ωR)2 +.5Iωf2.
Then I plugged .5MR2 in for I and πR for h (circumference divided by 2 for the center of the mass of the rope) This gave me:
mgπR+.25MR2ω02=.5m(ω0R)2 +.25MR2ωf2.
Then solving for ωf gave me:
√4πmgR+.25MR2ω02-.5m(ω0R)2/MR2

Does that look right?
 

tiny-tim

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mgπR+.25MR2ω02=.5m(ω0R)2 +.25MR2ωf2.
Ah, now i see what everything is! :smile:

ok, you've left out the final KE of the rope, and the initial KE of the rope is on the wrong side of the equation …

otherwise ok! :wink:
 
Re: Rotation

π²³ ∞ ° → ~ µ ρ σ τ ω ∑ … √ ∫ ≤ ≥ ± ∃ · θ φ ψ Ω α β γ δ ∂ ∆ ∇ ε λ Λ Γ ô


o right, that makes sense.

Okay, so now the equation is:
mgπR+.25MR2ω02+.5m(ω0R)2=.25MR2ωf2+.5m(ωfR)2

So now solving for ωf gives:
√mgπR+.25MR2ω02+.5m(ω0R)2/.25MR2+.5mR2

Does that look good? Thanks!
 
Re: Rotation

I went ahead and tried the second part of the question using that answer for ω and it was right. Thank you so much for your help!
 

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